a soccer player running on a level playing field kicks a soccer ball with a velocity of 15 m/s at an angle…

Explanation:

Step1: Find the vertical component of initial velocity

The initial velocity (v = 15\ m/s) and the angle (\theta=60^{\circ}). The vertical component (v_{0y}=v\sin\theta). [v_{0y}=15\sin60^{\circ}=15\times\frac{\sqrt{3}}{2}\approx12.99\ m/s]

Step2: Calculate the time of flight (a)

Using the kinematic equation (y = v_{0y}t-\frac{1}{2}gt^{2}). When the ball returns to the ground (y = 0). So (0=v_{0y}t-\frac{1}{2}gt^{2}), (t\left(v_{0y}-\frac{1}{2}gt\right)=0). One solution is (t = 0) (initial time), and the other is (t=\frac{2v_{0y}}{g}). Taking (g = 9.8\ m/s^{2}) [t=\frac{2\times15\sin60^{\circ}}{9.8}=\frac{2\times12.99}{9.8}\approx2.65\ s]

Step3: Calculate the range (b)

The horizontal component (v_{0x}=v\cos\theta). (v_{0x}=15\cos60^{\circ}=7.5\ m/s). Using the formula (R = v_{0x}t) [R=7.5\times2.65 = 19.875\ m]

Step4: Calculate the maximum height (c)

At maximum height, the vertical velocity (v_{y}=0). Using the kinematic equation (v_{y}^{2}-v_{0y}^{2}=-2gh). So (h=\frac{v_{0y}^{2}}{2g}) [h=\frac{(15\sin60^{\circ})^{2}}{2\times9.8}=\frac{12.99^{2}}{19.6}\approx8.63\ m]

Answer:

(a) (t\approx2.65\ s) (b) (R\approx19.88\ m) (c) (h\approx8.63\ m)